Three stars, each with the mass of our sun, form an equilateral triangle with sides 1.0 x 10¹² m long. (This triangle would just about fit within the orbit of Jupiter.) The triangle has to rotate, because otherwise the stars would crash together in the center. What is the period of rotation?

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Welcome back. Everyone in this problem, four celestial objects, each possessing a mass equivalent to the earth, arrange themselves into a square configuration which with each side measuring 1.4 multiplied by 10 to the 11th meters. In order to maintain this structure and prevent a gravitational collapse toward the center, the square must uniformly rotate. What is the duration of one complete rotation period? A says it's 3.15 multiplied by 10 to the 18th seconds. B 1.65 multiplied by 10 to the 15th seconds. C 3.47 multiplied by 10 to the 12 seconds and D one multiplied by 10 to the 10th seconds. Now, if we're going to figure out the duration of one complete rotation period, OK. And we're only given the length of each side for our square configuration, it means we have to understand our square configuration in order to solve this problem. So let's try to recreate it. So I'm gonna draw a square here. OK. And I'm gonna make the square as big as I can and at the corner of each set or at the corner of each or at each corner of the square, sorry, we're going to have our stars with a mass of M. We also know that the stars rotate about the center of mass, which is C as shown in our figure here. This is our center. And we can say that the distance from each star to the center of mass is R. Ok, let's put a distance here and let's call it R. Now, what lengths do we know? Well, we know that each side has a length of 1.4 multiplied by 10 to the 11th meters. So let's call that L. OK. And we also know that since this is a square, the angle between R and L is going to be 45 degrees. So that means we can say R equals L multiplied by the cosine of 45 degrees. OK. Now, if we think about it as well, when we think about the forces that are acting, we know that there's going to be a gravitational force acting on us from our celestial bodies. And we'll have two unique forces. First, we're going to have a gravitational force that's acting between adjacent stars. In other words, is the force and a star from the adjacent star. So let's call that gravitational force FG. We also know that there's going to be a gravitational force on a star from the diagonally opposite star which I can represent as or represent in blue highlighter here. And let's call that force FG prime OK. And we also know that since we have our adjacent forces FG, OK, then we're going to have the component of FG acting towards the S and or a component of FG acting towards the center as FG cost 45 in the same way that R is L cost 45. OK. Now, what do we know about our gravitational forces here? Well, recall, OK, that the gravitational force is equal to the gravitational constant multiplied by both masses. OK. Divided by the distance between the masses. In this case, all of our masses are the same. So M one and M two are equal to M. And thus, this tells us then that first, our gravitational force FG is going to be equal to GM squared divided by the distance between them. The, the, the distance between the adjacent star is L. So that's L squared. OK. And we also know that the distance between diagonal opposite stars equals GM squared divided by the distance between them. In this case, it's not given. But notice that this diagonal distance is the hypotenuse of a right triangle that's formed on one side of our square. So that means this distance equals the square root of L squared plus L squared or the square root of two L squared. And the square root of two L squared squared equals two L squared. So that gravitational force equals GM squared divided by two L squared. OK. So we know what our gravitational forces are, we also know that there's going to be a centripetal force that's between, that's between the center of mass and our celestial body. And it's the same for all of these four stars. So we could just represent it here as FC. OK. But bear in mind this force, this centripetal force is the same force that would be over here. No, the gravitational pull between all of these four stars. since they provide the necessary centripetal force, all of the stars are experiencing the same gravitational pull, since all of them are of equal masses and they are equally distant from each other. So if we're going to think about the rotation period, first, let's ask ourselves, what do we know about the centripetal force? Well, again, recall that our centripetal force is equal to MV squared divided by R and V or speed. OK, is equal to two pi R divided by T. So if we can find the central petal for us, OK. And put it into this equation. Substitute V in this equation, then we should be able to solve for our period. So let's do that. What else do we know about our centripetal force? Well, based on what we have here, we know that the centripetal force is going to be equal to those two components of our adjacent gravitational force. OK. So that's going to be two multiplied by FG cost 45 degrees plus the gravitational force between diagonally opposite styles FG prime. No, we already have expressions for FG and FG prime. So if we substitute, we get our centripetal force to be equal to two multiplied by GM squared divided by L squared plus 45 plus GM squared divided by two L squared. Now, since we have both of these or both of these expressions for our centripetal force. OK, then that means if we quit both of these, then our mass M OK? Multiplied by V squared, which we know is gonna be two pi R divided by T all squared divided by R all divided by R is going to be equal to two GM squared plus 45 degrees divided by L squared plus GM squared divided by two L squared. Now let's expand V squared. OK. So it's two pi R divided by tr squared. And when we do expand, OK, let me just make some space by taking it out here, then we're going to end up with four pi squared R squared divided by T squared. Now, if we look at both terms, OK, notice that we can cancel M from all, all our three terms M into M squared goes M times OK. We can also cancel R or R from our right or left hand side, R into R squared goes R times OK. And then we can factorize on our right hand side in terms of G two G and L squared. OK. So when we do that. Then on our left hand side, we're left with a four pi squared, R divided by T squared being equal to GM divided by L squared. OK. Multiplied by two costs 45 plus a half. OK? No, remember we're solving for T squared or we're solving for T because no, that means when we do that, if we solve for T, then T squared is going to be equal to uh T squared is gonna be equal to four pi squared. RL squared. OK. Divided by two cost 45 plus a half, which we can simplify to 1.914 multiplied by GM. So now we're almost there, we're almost there. But remember here, R we, we didn't figure R out but we know R equals L cost 45. So we can substitute R for L cost 45 here and find the spirit of both sides, which then tells us that T is gonna be equal to the square root of four pi squared multiplied by L cost 45 multiplied by L squared, which is gonna be LQ cost 45 degrees divided by 1.914 GM. Now we can solve for our PT because no, that tells us, OK, that is gonna be equal to four pi squared multiplied by 1.4 multiplied by 10 to the 11th meters cubed multiplied by the cosine of 45 degrees divided by 1.914 multiplied by gravitational constant G, 6.67 multiplied by 10 to the negative 11th Newton square meters per square kilogram. OK. Multiplied by the mass of the earth which we know is six multiplied by 10 to the 24th kilograms. So this is the expression we can use to solve for T and when we put that into our calculator, then to three significant figures, we get t to be approximately equal to one multiplied by 10 to the 10th seconds. Thus, that's the period, that's the period for, uh, of sorry, that's one complete rotational period. T. And if we look back at our answer choices, then that tells us D is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

Three stars, each with the mass of our sun, form an equilateral triangle with sides 1.0 x 10¹² m long. (This triangle would just about fit within the orbit of Jupiter.) The triangle has to rotate, because otherwise the stars would crash together in the center. What is the period of rotation?

Verified Solution

Key Concepts

Gravitational Force

Centripetal Force

Rotational Dynamics